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  4. Is there an error in this relativistic thought experiment?
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Is there an error in this relativistic thought experiment?

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Offline AndroidNeox (OP)

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Re: Is there an error in this relativistic thought experiment?
« Reply #20 on: 02/01/2019 17:01:15 »
Quote from: Halc on 15/11/2018 02:05:32
The gravitational potential between those two points is finite, but enough for escape speed to be c. 

No, this is wrong. The event horizon is defined as the point where the gravitational potential energy barrier is infinite and that is why the escape velocity is c. To accelerate a test mass to c requires infinite energy. It's not a finite energy barrier unless there's no event horizon.
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Re: Is there an error in this relativistic thought experiment?
« Reply #21 on: 02/01/2019 18:07:17 »
Quote from: Bored chemist on 15/11/2018 19:48:46
But not true for the observer falling in.
This is not correct. The falling observer measures finite time in their fall to the event horizon. That does not in any way imply that the fall ever ends. Time for the falling observer (even ignoring special relativistic effects) slows asymptotically approaching zero time passage as the observer approaches the event horizon. Just as the area under a decaying exponential curve is finite while the curve is infinitely long, the forever slowing observer never reaches zero.
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Re: Is there an error in this relativistic thought experiment?
« Reply #22 on: 02/01/2019 18:10:44 »
Quote from: Halc on 15/11/2018 02:05:32
The light does in fact get there, and beyond.  Rocks really do fall into black holes, with nothing unusual about the event.  No Hawking radiation observed by the rock for instance.  It just doesn't fall into the black hole in the frame of this distant observer is all.

As with all of Relativity, the observations of all observers are consistent. However, there is no possible way to make the infinite value of the time observed by any observer (not only distant) stationary with respect to the black hole match the finite time you suggest. If the falling object reaches the event horizon in finite time then that event must be observable outside of the event horizon within finite time. But, as you say, it is not observable.

The Shapiro delay is infinite. That means light cannot travel to an event horizon in finite time. Unless you are suggesting that Shapiro is wrong?
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Re: Is there an error in this relativistic thought experiment?
« Reply #23 on: 02/01/2019 21:07:27 »
Quote from: Halc on 02/01/2019 18:20:45
Energy is not conserved over different reference frames, so that property is being applied in an invalid manner here.

No, that's incorrect. All frames under consideration in this problem are stationary with respect to the black hole. All light is traveling only radially. Energy is conserved.
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Offline yor_on

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Re: Is there an error in this relativistic thought experiment?
« Reply #24 on: 07/01/2019 09:47:22 »
" I'm in the tiny minority of people who believe that there is NOTHING beyond the event horizon, that there is no BEYOND the horizon ... i.e., that the event horizon itself is the closest thing to a "center" that a black hole has. "

I've seen that definition too Mike and it makes sense. Once you passed a event horizon, if we define that as being inside the region from where light can't leave, it seems to me that there will be no real definition of what a center should be. You're already inside a region where physics breaks down.
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Offline Halc

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Re: Is there an error in this relativistic thought experiment?
« Reply #25 on: 07/01/2019 12:14:42 »
Red and blue shifting of light is a frame-dependent thing, not an actual thing light does.  You need to specify a frame, but your statements don't tell me which one to use.

Quote from: AndroidNeox on 03/01/2019 00:38:06
Maybe you should read the original post and work through the thought experiment.
The OP doesn't specify any frame.  It says these things:
Quote
The blueshift a light beam will undergo when traveling from any point in space to an event horizon is infinite.
...
By the time the front of the laser beam intersects the event horizon, the beam is infinitely blueshifted.
These are frame dependent claims.  I can look at light from a distant galaxy and the light is red shifted to one observer moving away from that galaxy, and blue shifted to another observer in the same place, but moving towards the galaxy.  The light is not actually shifted in either case, but appears red or blue relative to different frames.

Ditto for the black hole.  Light is finite red shifted for a falling observe at (or even beyond) the EH of a black hole.  For an accelerating observer looking up at a stationary distant light source, the light appears blue shifted due to the time dilation of the gravity well, and yes, this approaches infinite dilation as the point approaches the horizon, but at the horizon itself is no more a valid accelerating frame than is the inertial frame of a photon. Such a frame violates the principle of relativity.

Quote from: yor_on on 07/01/2019 05:47:59
The light from the lasers platform is indeed blue shifted, but only as measured at the Event horizon. Passing that EV it is part of the singularity and might be described as 'infinitely blue shifted', although that makes no sense from the position of making a measurement. Outside, or at, the EV it can't be infinitely blue shifted.

The light the platform possibly can measure on will be reflected light, and that, as it comes back from the vicinity of the black hole, must be red shifted. You always need to be clear about what frame of reference you use when setting up a thought experiment like this. 
And yet you don't specify one.  If I put a mirror stationary (relative to the platform) near a black hole, the observer on the platform will see his own reflected light coming back the same frequency as it left.  Not so if the mirror is falling.  But you didn't specify above, which is the gist of most of my prior post.
You can get arbitrarily high blue or red shift, but the only infinite shift I can think of is that from a laser falling into the black hole as seen from the platform outside.
« Last Edit: 03/02/2022 12:51:28 by Halc »
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Re: Is there an error in this relativistic thought experiment?
« Reply #26 on: 07/01/2019 13:11:18 »
What?

are you telling me that  " If I put a mirror stationary (relative to the platform) near a black hole, the observer on the platform will see his own reflected light coming back the same frequency as it left. "

Sorry
Fail
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Re: Is there an error in this relativistic thought experiment?
« Reply #27 on: 07/01/2019 13:13:33 »
A frequency is the number of occurrences of a repeating event per unit of time. That means that the platform watching the red shifted light also will find it to be at another frequency. Light is 'observer dependent'. meaning that you can see a light quanta as being constant, no matter if it seems to red or blue shift from another frame of reference.
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Re: Is there an error in this relativistic thought experiment?
« Reply #28 on: 07/01/2019 14:09:15 »
The point being, you have a good mind Halc, I enjoy reading you.
 Adapt and overcome.

you will get there.
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Re: Is there an error in this relativistic thought experiment?
« Reply #29 on: 08/01/2019 11:45:38 »
Want to show me the contradictions?
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Re: Is there an error in this relativistic thought experiment?
« Reply #30 on: 08/01/2019 19:51:31 »
Halc, okay, see how you think but when it comes to 'compressing' or 'stretching' a wave then that is a measurement of 'peaks and troughs' in time. If you think of cosmological red shift for example then that can be described by two ships passing/leaving each others perimeter as they exchange light signals. The signals will 'stretch' and so be come red shifted after they passed each other. It's the same here, as in the classical ambulance example with a siren sounding high pitched before meeting you, to then become lower pitched as it leaves you. What changes it is 'relative motion' and 'time'. There is no real 'loss' of those 'photons' (intrinsic) energy in either example, at least not as I see it, but thought of as waves their frequency change due to it.
=
The 'relative motion' I mentioned can also be referred to as a 'accelerating expansion of space' but it is essentially the same idea, then called cosmological red shift. What we're discussing is a equivalence but now using 'gravity' and 'time' instead.

( Exchange the bouncing for someone standing at the 'bottom of a gravity well'/event horizon sending a light signal to our platform. It must red shift. https://astronomy.swin.edu.au/cosmos/G/Gravitational+Redshift ). The point with me saying that I don't see it as a photon losing energy is due to the fact that this is a relation (observer dependencies), but that is a case of interpretation, you can choose the one there although I won't agree :)

spelling

Or maybe I will change my mind there? Damn, one reason why I don't want too is that no matter how you treat 'gravity' here that light is in a geodesic, it's not 'decelerating', it's just a equivalence. One way around it would be to define it as a 'field' in where there exist observer dependencies. Then define a 'photon' to be non propagating, but still behaving 'as if' to our measurements. That should mean that as the 'field strength' of 'gravity' diminish with distance the 'photon' would 'reclaim' its original energy.

Anyway, it's still observer dependent.
And gravity as a 'field'?
=

Here, read this and decide for yourself Halc :)
http://iopscience.iop.org/article/10.1088/1742-6596/600/1/012055/pdf

It's soo tricky that one. And btw, you can let a 'photon' propagate for that 'field' too if you like, reaching a same conclusion, but it's more to my taste treating it as a result of your measurement in time and space, no 'motion' necessary :)
« Last Edit: 08/01/2019 21:17:33 by yor_on »
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Re: Is there an error in this relativistic thought experiment?
« Reply #31 on: 09/01/2019 05:53:38 »
No, it was discussing the last part I wrote about, how to see the 'energy' contained in a 'photon' as it propagates.

"  Okun et al. note that in the literature there are two interpretations of the gravitational redshift in a static gravitational field: either the photon frequency is modified en route between emitter and receiver (and the clock rate unaffected by gravitational potential) or the clocks at lower potential are slowed down (and the photon unaffected en route).

They advocate strongly the clocks-slow-down view, stating that the gravitational redshift should be taught in a way that “centers on the universal modification of the rate of a clock exposed to a gravitational potential”. 

But the situation is subtle and confusing because an important heuristic principle in GR is that the local effects of gravity can always be eliminated with a coordinate transformation. This follows from a version of the principle of equivalence of gravity and inertia that asserts that inertial forces and gravitational forces are one and the same physical effect "

And that one is tricky, at least I think so. The one with it becoming a red shift when reflected back from a event horizon, not so much. And a cosmological red shift is due to a accelerating expansion of the vacuum (space). Turning it around it is a equivalence to those ships passing each other producing a Doppler redshift. The same effect is there, distance growing between ships/galaxies etc.
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Re: Is there an error in this relativistic thought experiment?
« Reply #32 on: 09/01/2019 06:14:41 »
No Halc, I'm not overthinking it. Relativity is simple as well as confusing, the more one learn the trickier it becomes :) And the red shift shown in this example is no different from the red shift you get sending a light signal from Earth to space. Clocks ticks differently depending on elevation in a gravitational potential and that affects the frequency of light. Infalling light blueshifts, light leaving a gravity well will be found to redshift.

" The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. "

There is one difference more btw between Doppler and Cosmological red shift. In the later case the redshift continue to grow as the light propagate as the universe constantly is expanding. In the first case the light sent doesn't :) But in reality, as all space is accelerating expanding, that difference seems of a lesser importance to me. Expressed otherwise: The difference between Doppler and Cosmological redshift is just a question over what distance you measure.

https://einstein.stanford.edu/content/relativity/a11859.html

« Last Edit: 09/01/2019 06:35:51 by yor_on »
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Re: Is there an error in this relativistic thought experiment?
« Reply #33 on: 09/01/2019 13:32:53 »
Hmm

The idea of different clock rates impending on wavelength and frequency is a easy idea to check, and it has been checked over and over again. But I won't spoonfeed, check it up yourself, It's not me you will need to convince of your interpretation btw, it's the physics you need to correct first, aka relativity.

that will be interesting.
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Re: Is there an error in this relativistic thought experiment?
« Reply #34 on: 09/01/2019 14:51:23 »
Look Halc, this is what you wrote. "   If I put a mirror stationary (relative to the platform) near a black hole, the observer on the platform will see his own reflected light coming back the same frequency as it left. "

That's what I reacted on, the rest of it is you not checking your sources to see if I'm correct or not. The lasers light will red shift as it is reflected from a gravity well (EV) back to the 'platform' with the laser. The red shift involves a slower frequency and a longer wave length, due to different time rates, as it 'climbs' the gravity well.

Check it
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Re: Is there an error in this relativistic thought experiment?
« Reply #35 on: 09/01/2019 17:10:37 »
Quote from: Halc on 07/01/2019 16:35:35
Correct again, but we're looking at the reflected beam from the frame of the platform from which it was sent, reflected by a mirror stationary relative to that platform.  Regardless of any gravity well in which the mirror might be, that is going to result in no red or blue shift as seen by our platform observer.

Exactly correct. Gravitational redshift is entirely reversible. Traveling into space of lower gravitational potential energy, the light is blueshifted. Then, when the light travels back to the platform, the gravitational blueshift is exactly reversed by the gravitational redshift because the light is returning to a point of the same gravitational potential energy as it started.

If the mirror is moving downward at a constant speed, there will be a Doppler redshift of the light beam but that will be constant throughout the experiment.
« Last Edit: 09/01/2019 17:16:58 by AndroidNeox »
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Offline AndroidNeox (OP)

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Re: Is there an error in this relativistic thought experiment?
« Reply #36 on: 09/01/2019 17:15:36 »
Quote from: Halc on 07/01/2019 12:14:42
If I put a mirror stationary (relative to the platform) near a black hole

How do you define, "near a black hole"? As the experiment shows, and the Shapiro Delay confirms, light cannot travel from any point in space to an event horizon in finite time. How are you defining two points, stationary WRT each other, to be "near" each other when light cannot travel from one to the other in finite time?

How are you defining distance?
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Re: Is there an error in this relativistic thought experiment?
« Reply #37 on: 09/01/2019 17:26:09 »
Quote from: Halc on 09/01/2019 15:40:32
As I keep repeating, all these effects are quite real, but irrelevant.

I think I see where you're getting confused. The blueshift is not irrelevant but the very crux of the argument.

If the light beam travels from the platform to the event horizon, the beam will be infinitely blueshifted. That means the downward beam will contain infinitely many wave cycles. That means, *before* the front of the light beam can reach the event horizon, the light source must generate an infinite sequence of light waves. This requires infinite time at the light source. This means, before the front of the light beam reaches the event horizon, infinite time must pass at the light source. That is a causal sequence and is therefore frame independent. There are no valid frames of reference from which the front of the beam can reach the event horizon before infinite time has passed at the platform.

And, because the location of the platform is irrelevant to the validity of the thought experiment, the result is true for all points in space.

I hope this clarifies it for you.
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Re: Is there an error in this relativistic thought experiment?
« Reply #38 on: 09/01/2019 17:31:17 »
I did this diagram to help make the thought experiment easier to visualize:
https://photos.app.goo.gl/2BqU6nZW85oToXw77
« Last Edit: 09/01/2019 17:33:18 by AndroidNeox »
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Offline Halc

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Re: Is there an error in this relativistic thought experiment?
« Reply #39 on: 09/01/2019 17:34:35 »
Quote from: AndroidNeox on 09/01/2019 17:15:36
Quote from: Halc on 07/01/2019 12:14:42
If I put a mirror stationary (relative to the platform) near a black hole

How do you define, "near a black hole"?
Outside the event horizon somewhere, but deeper in the gravity well than is the light source.

Quote
As the experiment shows, and the Shapiro Delay confirms, light cannot travel from any point in space to an event horizon in finite time. How are you defining two points, stationary WRT each other, to be "near" each other when light cannot travel from one to the other in finite time?
The light is not traveling to the event horizon.  It goes to the mirror and back.

Quote
How are you defining distance?
I didn't specify any distance.

Quote from: AndroidNeox on 09/01/2019 17:26:09
Quote from: Halc on 09/01/2019 15:40:32
As I keep repeating, all these effects are quite real, but irrelevant.

I think I see where you're getting confused. The blueshift is not irrelevant but the very crux of the argument.
You said earlier that the shift in the two directions cancel each other.  That makes it pretty irrelevant in this example.  I'm not claiming it doesn't happen, but I've not bothered to specifiy any distances or difference in potential since the light will be measured without redshift regardless of the values specified.  All I specified was that nothing was moving: the path is unchanged.

Quote
If the light beam travels from the platform to the event horizon, the beam will be infinitely blueshifted.
I never posted light going to the event horizon in my statement.  It doesn't come back if it does that.
I know you are the OP and may be imagining a different scenario, but the scenario I've been describing is reflecting light from a mirror somewhere outside the event horizon.

Quote from: yor_on on 09/01/2019 14:51:23
Look Halc, this is what you wrote. "   If I put a mirror stationary (relative to the platform) near a black hole, the observer on the platform will see his own reflected light coming back the same frequency as it left. "

That's what I reacted on, the rest of it is you not checking your sources to see if I'm correct or not. The lasers light will red shift as it is reflected from a gravity well (EV) back to the 'platform' with the laser. The red shift involves a slower frequency and a longer wave length, due to different time rates, as it 'climbs' the gravity well.

Check it
Not disputing that at all.  My comment above doesn't say otherwise.
But light does the opposite effect (negative redshift) when 'falling' into the gravity well in the first place.  Net effect from the platform is zero since it is at the same potential as the platform.  As I keep repeating, all these effects are quite real, but irrelevant.

I ran your assertion into a contradiction, so the analysis must be wrong according to you, but you don't point out where.
« Last Edit: 24/11/2021 01:37:29 by Halc »
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